a=superman is able to prevent evil
b= superman is impotent
c=superman is willing to prevent evil
d=superman is malevolent
e= superman exsists
f= superman prevents evil
(we have to prove ¬e)
- (a∧c)→f =”p”→f =add !f = apply modus tollens (!f∧(p→f))→!p = !p = !(a∧c)
- ¬ (a∧c) = !a V !c
*i am lost after this
i know this is supposed to look like this
and pg 73 of the book has it but i dont really understand how the substituting works
this website explains more about building arguments:
T(a,b) = Student a likes TV show b where the domain for a is the students in the class and the domain for b is the shows currently airing on TV
T(s,u) = all students s like a TV show u
or everybody likes a TV show
¬T(t,u) = it is not the case that a student t likes the TV show u
its not true that one student likes the TV show
if everybody likes a TV show, then its not true that one student likes the TV show,
* can this be translated into :
if everybody likes a TV show, then there is not one student who doesnt like the TV show?
use quantifiers and variable to translate the following prop
there is a TV show that everybody likes.
on the test i put this:
*should it be
4a. Prove that if x is a rational number and x ≠ 0 then 1/x is a rational number
If x is a rational number ≠ 0, then it can be written as a/b, where a and b are integers and ≠ 0. Then 1/x is equal to 1/(a/b) which equals b/a. Therefore if x is a rational number, 1/x is also rational.
4b. Prove that if x is an irrational number and x≠ 0 then 1/x is an irrational number.
Assume that x is irrational and 1/x is rational. Then 1/x can be written as a/b where a and b are integers ≠ 0. So x is equal to 1/(a/b) which equals b/a, b and a are integers. Therefore x is rational. This contradicts the assumption that x is irrational. Therefore it is true that if x is irrational, 1/x is also irrational.
6a. Determine the argument form of the above argument.
Notation: W(x): x is willing to prevent evil
A(x): x is able to prevent evil
P(x): x prevents evil
I(x): x is impotent
M(x): x is malevolent
E(x): x exists.
S stands for Superman.
Let’s write the givens:
1: (A(S) ∧ W(S)) → P(S)
2. ¬A(S) → I(S)
3. ¬W(S) → M(S)
5. E(S) → (¬M(S) ∧ ¬I(S))
6. ¬A(S) ∨ ¬W(S)
7. M(S) ∨ I(S)
8. ¬(¬M(S) ∧ ¬I(S))
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As announced in class on Thursday:
- This Thursday’s quiz will cover material/homework from Sections 2.4 and 2.5.
- Test #2 will be given in class on Thursday, April 6.
I hope everyone enjoyed yesterday’s snow day! We will still have Quiz #4 tomorrow on material/homework from Sections 2.3 and 2.4. Since I didn’t hold my usual office hours yesterday, I *might* be able to squeeze some in tomorrow at noon. Send me an email or comment on this post if you would like to see me then.
As announced in class yesterday, next week’s quiz will cover material/homework from Sections 2.3 and 2.4.